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Re: Relevant problem. How do you solve for y?

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  • video_ranger
    ... This is equivalent to: (y+a)^n + (y+b)^n = c Making various substitutions this is also equivalent to solving: (x+1)^n + (x-1)^n = K for x in terms of K. I
    Message 1 of 1 , Apr 17, 2013
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      --- In mathforfun@yahoogroups.com, bqllpd <no_reply@...> wrote:
      >
      > \sum_{i=0}^n y^{n-i}*\binom{n}{i}*(a^i+b^i)=c Solve for y.
      >

      This is equivalent to:

      (y+a)^n + (y+b)^n = c

      Making various substitutions this is also equivalent to solving:

      (x+1)^n + (x-1)^n = K

      for x in terms of K. I don't think there's any formula except for certain values of n. I tried asking Wolfram alpha to solve

      (x+1)^5 + (x-1)^5 = K

      for x and there was no formula in the usual sense.

      I think your posts will have to be moderated again.

      Mark
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